Chapter 3 of 13

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Andhra Pradesh (BIEAP)Class 9 Mathematics★ AP Board (BSEAP) Class 9 Mathematics · ~10 marks · Board: AP Board

Polynomials and Factorisation

POLYNOMIALS — degree, types (linear, quadratic, cubic). ZEROES of a polynomial. REMAINDER THEOREM and FACTOR THEOREM. Algebraic IDENTITIES. Factorising CUBIC polynomials. Complete AP Class 9 Mathematics.

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By the end of this chapter you'll be able to…

1Classify polynomials by degree (constant, linear, quadratic, cubic) and identify zeroes
2Apply Remainder Theorem: find remainder when p(x) is divided by (x−a) by evaluating p(a)
3Apply Factor Theorem: verify that (x−a) is a factor when p(a)=0
4Factorise cubic polynomials: (1) find one zero by trial, (2) divide, (3) factorise quadratic
5Use all 8 algebraic identities including the cubic and (a+b+c)² identities
6Evaluate numerical expressions using identities (avoid direct multiplication)
7Solve problems where a coefficient in a polynomial is unknown (find k using Remainder Theorem)

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Why this chapter matters

Polynomials is among the highest-scoring algebra chapters in AP Class 9 Mathematics. The Remainder Theorem and Factor Theorem together generate 4-5 mark problems. Factorising cubic polynomials using the Factor Theorem is a standard 4-5 mark question. Identity-based evaluation questions (evaluate 105×95 using a²−b²) appear for 2-3 marks. The cubic identities (a³+b³, a³−b³) are tested every year. A strong foundation here directly supports Class 10 Algebra (quadratic equations, algebraic fractions).

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Polynomials and Factorisation — Class 9 Mathematics

1. Polynomials — Definition and Types

p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ (aₙ ≠ 0). Degree = n (highest power). Types by degree: Constant (degree 0). Linear (degree 1). Quadratic (2). Cubic (3).

2. Zeroes of a Polynomial

A real number 'a' is a ZERO of p(x) if p(a) = 0. A linear polynomial has EXACTLY 1 zero. A quadratic polynomial can have at MOST 2 zeroes. Geometry: zero = x-coordinate where graph crosses x-axis.

3. Remainder Theorem

When p(x) is divided by (x−a), the REMAINDER = p(a). 'You don't need to perform long division to find the remainder. Just PLUG x=a into the polynomial.'

Example: Find remainder when p(x)=x³−2x²+4x−5 is divided by (x−2). p(2) = 8−8+8−5 = 3. The remainder is 3.

4. Factor Theorem

(x−a) is a FACTOR of p(x) ⇔ p(a) = 0. The converse of the Remainder Theorem. Example: Is (x+1) a factor of x³+x²−x−1? p(−1) = −1+1+1−1 = 0. YES — (x+1) is a factor.

Factorising Cubic Polynomials

Find ONE zero by trial (check factors of constant term). 2. Divide the cubic by (x−zero) to get a quadratic. 3. Factorise the quadratic (by splitting middle term).

5. Key Algebraic Identities

Identity	Use
(a+b)² = a²+2ab+b²	Expand
(a−b)² = a²−2ab+b²	Expand
a²−b² = (a+b)(a−b)	Factorise difference of squares
(x+a)(x+b) = x²+(a+b)x+ab	Expand
a³+b³ = (a+b)(a²−ab+b²)	Factorise sum of cubes
a³−b³ = (a−b)(a²+ab+b²)	Factorise difference of cubes
(a+b)³ = a³+3a²b+3ab²+b³	Expand
(a+b+c)² = a²+b²+c²+2ab+2bc+2ca	Expand

6. Worked Example — Factorising a Cubic

Factorise: x³ − 3x² − 9x − 5. Try factors of constant (−5): ±1, ±5. p(−1) = −1−3+9−5 = 0. So (x+1) is a factor. Divide: (x³−3x²−9x−5) ÷ (x+1) = x²−4x−5. Factorise quadratic: x²−4x−5 = (x−5)(x+1). Full factorisation: x³−3x²−9x−5 = (x+1)(x+1)(x−5) = (x+1)²(x−5).

Common Mistakes

Confusing (a+b)³ EXPANSION with a³+b³ FACTORISATION. They are DIFFERENT.
Forgetting that (a+b)³ has FOUR terms (a³+3a²b+3ab²+b³), not two.

More Worked Examples

Example — Finding Zeroes Graphically: Find the zeroes of p(x) = x²−2x−8. Factorise: (x−4)(x+2). Zeroes: 4, −2. Check: p(4) = 16−8−8 = 0 ✓. p(−2) = 4+4−8 = 0 ✓.

Example — Factor Theorem Application: Is (x+3) a factor of p(x) = x³+x²−9x−9? p(−3) = −27+9+27−9 = 0. YES. Factor: (x+3)(x²−2x−3) = (x+3)(x−3)(x+1).

Example — Identity Application: Evaluate 105×95 using identity. 105×95 = (100+5)(100−5) = 100²−5² = 10000−25 = 9975.

Key Exam Tips

For Remainder Theorem: the remainder when dividing by (x−a) is p(a). For (x+a), evaluate p(−a).
For factorisation: FIRST check for common factors. THEN try identities. THEN split middle term. THEN Factor Theorem for cubics.
'ALWAYS verify your factorisation by expanding back — it takes 10 seconds and catches errors.'

Types of Polynomials by Degree

Constant (degree 0): p(x) = 5. No variable. Graph: horizontal line. Linear (degree 1): p(x) = 2x+3. ONE zero. Graph: straight line. Quadratic (degree 2): p(x) = x²−4x+3. UP TO 2 zeroes. Graph: PARABOLA. Cubic (degree 3): p(x) = x³−6x²+11x−6. UP TO 3 zeroes. Graph: S-curve.

Graph of a Quadratic Polynomial

y = ax²+bx+c is a PARABOLA. a > 0: opens UPWARD (∪ — minimum). a < 0: opens DOWNWARD (∩ — maximum). The zeroes are the x-coordinates where the parabola crosses the x-axis. 'The number of zeroes = the number of times the graph crosses the x-axis. This is the GEOMETRIC MEANING of zeroes.'

Quick Revision — Key Identities

(a+b)² = a²+2ab+b². (a−b)² = a²−2ab+b². a²−b² = (a+b)(a−b). (x+a)(x+b) = x²+(a+b)x+ab. (a+b+c)² = a²+b²+c²+2ab+2bc+2ca. a³+b³ = (a+b)(a²−ab+b²). a³−b³ = (a−b)(a²+ab+b²).

Factorisation Strategy — Step by Step

When asked to factorise a polynomial, follow this order:

Check for a common factor — if every term has a common factor, take it out FIRST. Example: 3x³+6x²+3x = 3x(x²+2x+1) = 3x(x+1)².
Check if it fits an identity — a²−b², a²+2ab+b², a³±b³, etc.
Split the middle term (quadratics) — x²+5x+6. Find two numbers with sum=5, product=6 → 2 and 3 → (x+2)(x+3).
Factor Theorem (cubics) — find one zero by trial, divide, factorise the quadratic.

Common Factorisation Mistakes and Fixes

Mistake	Why It Happens	Fix
x²−4 = (x−2)²	Confusing a²−b² with (a−b)²	a²−b² = (a+b)(a−b). (a−b)² = a²−2ab+b². They are DIFFERENT.
x³−8 = (x−2)(x²+2x+4) but student writes x²+4	Forgetting the middle term in a³−b³	Formula: a³−b³ = (a−b)(a²+ab+b²). The "ab" term is needed.
Forgetting common factor first	Jumping to complex methods	ALWAYS check for common factors before anything else.
Sign errors in splitting middle term	Rushing	Verify: sum and product must be EXACT. Check sign of constant term.

Advanced Worked Examples

Example — Factorise 4x² − 25y²: This is a²−b² with a=2x, b=5y. 4x²−25y² = (2x+5y)(2x−5y).

Example — Factorise x⁴ − 16: Treat as (x²)² − 4² = (x²+4)(x²−4) = (x²+4)(x+2)(x−2). 'Difference of squares applied TWICE — a common 3-mark question.'

Example — Evaluate Using Identity: 103²: 103² = (100+3)² = 100²+2(100)(3)+3² = 10000+600+9 = 10609.

Example — Factorise x³ − 23x² + 142x − 120: Try x=1 → 1−23+142−120 = 0. (x−1) is a factor. Divide: (x³−23x²+142x−120)÷(x−1) = x²−22x+120. Factorise quadratic: (x−10)(x−12). Full factorisation: (x−1)(x−10)(x−12). Zeroes: 1, 10, 12.

Example — Factorise 8x³ + 27y³: This is (2x)³+(3y)³ = (2x+3y)((2x)²−(2x)(3y)+(3y)²) = (2x+3y)(4x²−6xy+9y²).

Remainder Theorem — Word Problems

Problem: When p(x) = x³ + kx² − 4x + 2 is divided by (x−2), the remainder is 14. Find k. p(2) = 8 + 4k − 8 + 2 = 14 → 4k + 2 = 14 → 4k = 12 → k = 3.

Problem: p(x) = 2x³ + ax² + bx − 10. (x+1) gives remainder −18. (x−1) gives remainder −6. Find a and b. p(−1) = −2 + a − b − 10 = −18 → a − b = −6 ... (1). p(1) = 2 + a + b − 10 = −6 → a + b = 2 ... (2). Add (1) and (2): 2a = −4 → a = −2. From (2): −2 + b = 2 → b = 4. So a = −2, b = 4.

Division Algorithm for Polynomials

p(x) = g(x) × q(x) + r(x), where deg(r) < deg(g). Dividend = Divisor × Quotient + Remainder. 'Use this to VERIFY your division. If dividend ≠ divisor × quotient + remainder, you made an error in division.'

Exam Focus — AP Board (BSEAP)

Question Type	Marks
Identify degree/type of polynomial	1
Find zeroes (linear and quadratic)	2-3
Remainder Theorem application	2-3
Factor Theorem — check if (x−a) is a factor	2
Factorise cubic polynomial	4-5
Evaluate using identities (without direct calculation)	2-3

Common AP Exam Question Patterns

'Find the remainder when p(x) is divided by (x−a).' — Direct Remainder Theorem. Substitute x=a.
'Is (x+2) a factor of p(x)?' — Check if p(−2) = 0.
'Factorise x³ − 6x² + 11x − 6.' — Use Factor Theorem, find first zero by trial.
'Evaluate 102 × 98 using an identity.' — Write as (100+2)(100−2) = 100²−2².
'If (x−1) is a factor of x³+kx²−x−1, find k.' — p(1)=0, solve for k.

Key Exam Tips

'Without actual division' in the question means USE the Remainder Theorem.
'Using suitable identity, evaluate...' means do NOT multiply directly. Rewrite as a²−b² or (a±b)².
For cubic factorisation: ALWAYS try factors of the constant term first (±1, ±2, ±3, etc.).
After factorising, write 'Zeroes are...' and list them — many students forget this 1-mark step.
Memorise a³+b³ and a³−b³ identities — they appear in at least one question every year.

Quick Self-Test

Degree of polynomial p(x) = 4x⁵ − 3x² + 7? (Answer: 5.)
Zero of p(x) = 2x − 6? (Answer: x = 3.)
Remainder when x³+x²−3x+5 is divided by (x+1)? (Answer: p(−1)=−1+1+3+5=8.)
Factorise x² − 9. (Answer: (x+3)(x−3).)
Is (x−2) a factor of x³ − 8? (Answer: p(2)=8−8=0. Yes.)
Expand (x+3)³ using identity. (Answer: x³+9x²+27x+27.)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Polynomials, Theorems and Identities

POLYNOMIAL: p(x) = aₙxⁿ + ... + a₁x + a₀ (aₙ ≠ 0). DEGREE = highest power. Types: Constant (0), Linear (1), Quadratic (2), Cubic (3). ZERO: real number 'a' is a zero if p(a) = 0. Linear → exactly 1 zero. Quadratic → at most 2 zeroes. Cubic → at most 3 zeroes. REMAINDER THEOREM: When p(x) is divided by (x−a), REMAINDER = p(a). For (x+a), substitute x = −a. FACTOR THEOREM: (x−a) is a FACTOR of p(x) ⟺ p(a) = 0. FACTORISING CUBICS: Step 1: Find one zero by trial — try factors of constant term (±1, ±2, etc.). Step 2: Divide p(x) by (x − zero) → get quadratic. Step 3: Factorise the quadratic by splitting middle term. ALGEBRAIC IDENTITIES: (a+b)² = a²+2ab+b². (a−b)² = a²−2ab+b². a²−b² = (a+b)(a−b). (x+a)(x+b) = x²+(a+b)x+ab. (a+b)³ = a³+3a²b+3ab²+b³. (a−b)³ = a³−3a²b+3ab²−b³. a³+b³ = (a+b)(a²−ab+b²). a³−b³ = (a−b)(a²+ab+b²). (a+b+c)² = a²+b²+c²+2ab+2bc+2ca. FINDING UNKNOWN COEFFICIENTS: If remainder when dividing by (x−a) is R, then p(a) = R. Set up equation and solve for unknown. DIVISION ALGORITHM: Dividend = Divisor × Quotient + Remainder. WORKED EXAMPLES: Remainder of x³−2x²+4x−5 by (x−2): p(2) = 8−8+8−5 = 3. Factorise x³−3x²−9x−5: p(−1)=0 → (x+1) is a factor. Divide → (x+1)(x²−4x−5) = (x+1)(x+1)(x−5) = (x+1)²(x−5). Evaluate 105×95 = (100+5)(100−5) = 100²−5² = 9975.

AP EXAM KEY TRAPS: (1) For (x+a), evaluate p(−a) NOT p(a). e.g., dividing by (x+3) → substitute x = −3. (2) a³+b³ has MIDDLE TERM: (a+b)(a²−ab+b²). NOT (a+b)(a²+b²). The 'ab' term is always NEGATIVE in the bracket. Similarly a³−b³ has +ab in bracket. (3) (a+b)³ has FOUR TERMS: a³+3a²b+3ab²+b³. NOT a³+b³. (4) ALWAYS check: if (x−a) is claimed as a factor, verify p(a)=0. (5) FACTORISATION STRATEGY: common factor first → identities → split middle term → Factor Theorem for cubics. NUMERICAL IDENTITY TRICK: When asked to evaluate without direct multiplication, look for patterns: 102×98 = (100+2)(100−2) = 10000−4 = 9996.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing a³+b³ = (a+b)(a²−ab+b²) with expanding as (a+b)³

✓ THESE ARE DIFFERENT: (a+b)³ = a³ + 3a²b + 3ab² + b³ (expansion of a CUBE). It has FOUR terms with coefficients 1, 3, 3, 1. a³ + b³ = (a+b)(a²−ab+b²) (FACTORISATION of sum of cubes). Notice: (a+b)³ ≠ a³+b³. They are NOT the same thing. Similarly: (a−b)³ = a³ − 3a²b + 3ab² − b³. a³ − b³ = (a−b)(a²+ab+b²). KEY PATTERN for the sum/difference of cubes: a³ ± b³ = (a ± b)(a² ∓ ab + b²). The signs alternate: if the original is +, the bracket has −ab. If the original is −, the bracket has +ab. QUICK TEST: expand (a+b)(a²−ab+b²) = a³−a²b+ab²+a²b−ab²+b³ = a³+b³ ✓.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Polynomials and Factorisation?

1 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

1 questions~2 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•POLYNOMIAL: An algebraic expression of the form p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, where coefficients are real numbers and exponents are non-negative integers. DEGREE = highest power of x.
•TYPES BY DEGREE: Constant (degree 0, e.g., 5). Linear (degree 1, e.g., 2x+3). Quadratic (degree 2, e.g., x²−5x+6). Cubic (degree 3, e.g., x³−7x+6). Biquadratic (degree 4).
•ZEROES OF A POLYNOMIAL: A value 'a' is a zero of p(x) if p(a) = 0. Linear has exactly 1 zero. Quadratic has at most 2 zeroes. Cubic has at most 3 zeroes. The zeroes are also called ROOTS.
•REMAINDER THEOREM: When polynomial p(x) is divided by (x−a), the REMAINDER is p(a). For divisor (x+a), evaluate p(−a). This avoids long polynomial division for finding remainders.
•FACTOR THEOREM: (x−a) is a FACTOR of p(x) if and only if p(a) = 0. Used to verify factors and to find zeroes. If p(a) = 0, you can divide p(x) by (x−a) to get a polynomial of one less degree.
•FACTORISING CUBIC POLYNOMIALS (3-STEP METHOD): (1) Find one zero by trial — test factors of the constant term (±1, ±2, ±3, etc.). The zero 'a' gives factor (x−a). (2) DIVIDE p(x) by (x−a) using long division or synthetic division — get a quadratic. (3) Factorise the quadratic by splitting middle term.
•ALGEBRAIC IDENTITIES — QUADRATIC: (a+b)² = a²+2ab+b². (a−b)² = a²−2ab+b². a²−b² = (a+b)(a−b). (x+a)(x+b) = x²+(a+b)x+ab.
•ALGEBRAIC IDENTITIES — CUBIC: (a+b)³ = a³+3a²b+3ab²+b³. (a−b)³ = a³−3a²b+3ab²−b³. a³+b³ = (a+b)(a²−ab+b²). a³−b³ = (a−b)(a²+ab+b²). Note the alternating sign in the bracket (−ab for sum of cubes, +ab for difference).
•TRINOMIAL EXPANSION: (a+b+c)² = a²+b²+c²+2ab+2bc+2ca. (a+b+c)³ has 10 terms — usually not memorised; expand step by step.
•NUMERICAL EVALUATION USING IDENTITIES: 102×98 = (100+2)(100−2) = 100²−2² = 10000−4 = 9996. 103² = (100+3)² = 10000+600+9 = 10609. 99×101 = (100−1)(100+1) = 10000−1 = 9999. Look for patterns instead of multiplying directly.
•FINDING UNKNOWN COEFFICIENT: If the remainder when p(x) is divided by (x−a) is R, then p(a) = R. Substitute, solve for the unknown. Example: find k if x³+kx²−4x+2 ÷ (x−2) has remainder 14. p(2) = 8+4k−8+2 = 4k+2. Set 4k+2 = 14 → k = 3.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Engineering and polynomial modelling

Engineers model real-world phenomena using polynomials. The trajectory of a projectile (cricket ball, rocket) follows a quadratic curve y = ax² + bx + c. The bending of a beam under load follows a polynomial. Stress-strain curves in materials science are polynomial. Polynomial root-finding (factor theorem) is fundamental to solving these engineering equations. AP's growing manufacturing sector (Vizag Steel, AP Industries) employs engineers who use polynomial mathematics daily.

Computer graphics and animation

Bezier curves (used in fonts, animation, vector graphics — every letter in this text was drawn with Bezier curves) are polynomial curves. A quadratic Bezier uses degree-2 polynomials; cubic Beziers use degree-3 polynomials. Animation interpolation (smoothly moving an object between keyframes) uses polynomial functions. AP's animation and gaming industry (Hyderabad's T-Hub, growing studios) employs developers who work with polynomial mathematics for visual rendering.

Coding theory and error correction

Modern data storage (DVDs, hard drives, cloud storage) uses error-correcting codes based on POLYNOMIALS — specifically, Reed-Solomon codes. The data is encoded as the coefficients of a polynomial; if some data bits are corrupted, the original polynomial can be reconstructed using the surviving values. The Factor Theorem and polynomial evaluation underlie this entire field. Every QR code, mobile signal, and satellite communication uses polynomial-based error correction — a direct application of Class 9 algebra.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Remainder Theorem (3-4 marks): identify the divisor (x−a) or (x+a). State 'By Remainder Theorem, remainder = p(a) [or p(−a)].' Substitute and compute. Three explicit steps = full marks.
Factor Theorem verification (2-3 marks): substitute x=a into p(x). If p(a)=0, state 'Therefore (x−a) is a factor.' If p(a)≠0, state 'Since p(a)≠0, (x−a) is NOT a factor.' Always conclude explicitly.
Cubic factorisation (4-5 marks): write the THREE-STEP method clearly. (1) Try factors of constant, find one zero. (2) Divide to get quadratic. (3) Factorise quadratic. Each step should be shown explicitly. The final answer should be written as a product of linear factors.
Identity-based evaluation (2-3 marks): write the substitution explicitly. '102 × 98 = (100+2)(100−2) [using a²−b² identity] = 100² − 2² = 10000 − 4 = 9996.' Showing the identity used earns the method mark.
Finding unknown coefficient (3-4 marks): apply Remainder Theorem: p(a) = remainder. Substitute the unknown coefficient k. Set up the equation, solve for k. Write 'k = ___' as the final answer. Don't forget to verify by re-substituting.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research the Fundamental Theorem of Algebra (Gauss, 1799) — every polynomial of degree n has exactly n complex roots (counting multiplicity). For real polynomials, complex roots come in conjugate pairs. This is one of the most important theorems in algebra and underlies all polynomial theory. Research how Gauss proved it (he gave 4 different proofs over his lifetime).
Investigate Galois Theory (Évariste Galois, 1832) — Galois proved that polynomials of degree 5 or higher cannot be solved by a general formula (no 'quintic formula' analogous to the quadratic formula). This was discovered when Galois was 18 years old. He died in a duel at 20. His theory founded modern abstract algebra. Research the connection between polynomial roots and group theory.
Explore symmetric polynomials and Vieta's formulas — for a polynomial with roots r₁, r₂, ..., rₙ, the SUM of roots = −(coefficient of xⁿ⁻¹)/(leading coefficient), and the PRODUCT of roots = (−1)ⁿ × (constant term)/(leading coefficient). For x²+bx+c with roots p, q: p+q=−b, pq=c. Research how these formulas extend to cubics and quartics — important in olympiad mathematics.
Research polynomial interpolation — given n+1 distinct data points, there is a UNIQUE polynomial of degree at most n that passes through all of them (Lagrange interpolation). This is used in numerical analysis, computer graphics, and data fitting. Research how this connects to the Factor Theorem (a polynomial of degree n with n+1 known values is uniquely determined).

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC (Class 10) — PolynomialsVery High — Class 9 Polynomials directly extends to Class 10 Polynomials (zeroes, division algorithm, relationship between zeroes and coefficients) and Quadratic Equations

JEE Main and AdvancedVery High — polynomials, factorisation, and algebraic identities form a major part of JEE Algebra

NTSE (Mathematics)High — polynomial questions appear in NTSE Stage I and II

Mathematics Olympiad (RMO, INMO)Very High — polynomial techniques (root finding, factorisation, symmetric functions) are essential for olympiad mathematics

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Questions students ask

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DIVISION ALGORITHM: For any polynomial p(x) and divisor d(x), we can write p(x) = d(x) × q(x) + r(x), where q(x) is the quotient and r(x) is the remainder. When the divisor is LINEAR (x−a), the remainder must be a CONSTANT (degree less than 1). So: p(x) = (x−a) × q(x) + R, where R is a constant. To find R, substitute x = a: p(a) = (a−a) × q(a) + R = 0 + R = R. Therefore R = p(a). This is why the Remainder Theorem works: substituting the value that makes the divisor zero gives the constant remainder. This single insight saves enormous amounts of polynomial long division.

RATIONAL ROOT THEOREM (simplified for Class 9): if a polynomial with integer coefficients has a rational zero p/q (in lowest form), then p must divide the CONSTANT TERM and q must divide the LEADING COEFFICIENT. For monic polynomials (leading coefficient 1), the rational zeroes must be integer DIVISORS of the constant term. EXAMPLE: x³−6x²+11x−6. Constant = −6. Divisors: ±1, ±2, ±3, ±6. Try these one by one: p(1) = 1−6+11−6 = 0 ✓. Found! Use (x−1) as a factor. STRATEGY: try ±1 first (easy mental calculation), then ±2, ±3. If the constant term is large, divisors are limited to relatively small numbers.

These are FUNDAMENTALLY DIFFERENT expressions. (a+b)³ means the cube of (a+b) — multiply (a+b) by itself three times. Expansion: (a+b)³ = a³ + 3a²b + 3ab² + b³ — FOUR TERMS with coefficients 1, 3, 3, 1. a³+b³ means just the sum of two cubes — TWO TERMS. Its FACTORISATION: a³+b³ = (a+b)(a²−ab+b²). NUMERIC CHECK: Let a=2, b=3. (a+b)³ = 5³ = 125. a³+b³ = 8+27 = 35. Clearly 125 ≠ 35. The difference: 3a²b + 3ab² = 3ab(a+b) = 3(6)(5) = 90, and 35 + 90 = 125 ✓. So (a+b)³ = a³+b³ + 3ab(a+b). Common error: confusing these. Always check by simple substitution.

EXPAND IT: (a+b)(a²−ab+b²) = a×a² + a×(−ab) + a×b² + b×a² + b×(−ab) + b×b² = a³ − a²b + ab² + a²b − ab² + b³. The middle terms cancel: −a²b + a²b = 0 and ab² − ab² = 0. Left with: a³ + b³. Therefore (a+b)(a²−ab+b²) = a³+b³. Similarly, (a−b)(a²+ab+b²) = a³−b³ (expand to verify). MEMORY RULE: For sum of cubes a³+b³, the factor is (a+b) and the bracket has SAME sign (a²) but OPPOSITE middle sign (−ab) and POSITIVE last (b²). For difference of cubes a³−b³, the factor is (a−b) with bracket signs opposite: (a²+ab+b²).

Recognise the pattern: 998 = 1000−2 and 1002 = 1000+2. So 998 × 1002 = (1000−2)(1000+2) = 1000² − 2² (using a²−b² identity) = 1000000 − 4 = 999996. The identity (a+b)(a−b) = a²−b² turns a multiplication into a subtraction of squares — vastly easier. SIMILAR EXAMPLES: 25 × 35 = (30−5)(30+5) = 900−25 = 875. 96 × 104 = (100−4)(100+4) = 10000−16 = 9984. 47² = (50−3)² = 2500 − 300 + 9 = 2209. The key skill: SEE THE PATTERN — if two numbers are equidistant from a round number (like 100, 50, 1000), use the difference of squares trick.

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